Resolved shear stress

6.3.2 Resolved shear stress

All working processes such as rolling, extrusion, forging, etc. cause plastic deformation and, con- sequently, these operations will involve the processes of slip or twinning outlined above. The stress system applied during these working operations is often quite complex, but for plastic deformation to occur the presence of a shear stress is essential. The importance of shear stresses becomes clear when it is realized that these stresses arise in most processes and tests even when the applied stress itself is not a pure shear stress. This may be illustrated by examining a cylindrical crystal of area A in

a conventional tensile test under a uniaxial load P. In such a test, slip occurs on the slip plane, shown shaded in Figure 6.9, the area of which is A/cos φ, where φ is the angle between the normal to the plane OH and the axis of tension. The applied force P is spread over this plane and may be resolved into a force normal to the plane along OH, Pcos φ, and a force along OS, Psin φ. Here, OS is the line of greatest slope in the slip plane and the force Psin φ is a shear force. It follows that the applied

stress (force/area) is made up of two stresses, a normal stress (P/A)cos 2 φ tending to pull the atoms apart, and a shear stress (P/A) cos φ sin φ trying to slide the atoms over each other. In general, slip does not take place down the line of greatest slope unless this happens to coincide with the crystallographic slip of direction. It is necessary, therefore, to know the resolved shear stress

Mechanical properties I 299 on the slip plane and in the slip direction. Now, if OT is taken to represent the slip direction, the

resolved shear stress will be given by σ = Pcos φ sin φ cos χ/A, where χ is the angle between OS and OT. Usually this formula is written more simply as

σ = Pcos φ cos λ/A, (6.4) where λ is the angle between the slip direction OT and the axis of tension. It can be seen that the

resolved shear stress has a maximum value when the slip plane is inclined at 45 ◦ to the tensile axis, and becomes smaller for angles either greater than or less than 45 ◦ . When the slip plane becomes more nearly perpendicular to the tensile axis (φ > 45 ◦ ), it is easy to imagine that the applied stress has a greater tendency to pull the atoms apart than to slide them. When the slip plane becomes more nearly parallel to the tensile axis (φ < 45 ◦ ), the shear stress is again small but in this case it is because the area of the slip plane, A/cos φ, is correspondingly large.

A consideration of the tensile test in this way shows that it is shear stresses which lead to plastic deformation, and for this reason the mechanical behavior exhibited by a material will depend, to some extent, on the type of test applied. For example, a ductile material can be fractured without displaying its plastic properties if tested in a state of hydrostatic or triaxial tension, since under these conditions the resolved shear stress on any plane is zero. Conversely, materials which normally exhibit

a tendency to brittle behavior in a tensile test will show ductility if tested under conditions of high shear stresses and low tension stresses. In commercial practice, extrusion approximates closely to a system of high shear stress, and it is common for normally brittle materials to exhibit some ductility when deformed in this way (e.g. when extruded).

Worked example

A single crystal of iron is pulled along [1 2 3]. Which is the first slip system to operate?

Solution

Slip plane

Cos φ

Burgers vector

Cos λ

Schmid factor √ √ × 28 × 42

4/ 42 −4 Slip will occur on (1 0 1)[1 ¯1 ¯1] first.

300 Physical Metallurgy and Advanced Materials